Chứng minh rằng: $\frac{2sinxcosx-sin4x}{2.cos^22x-cos2x}$ =tan(-2x) 14/10/2021 Bởi Anna Chứng minh rằng: $\frac{2sinxcosx-sin4x}{2.cos^22x-cos2x}$ =tan(-2x)
$VT=\dfrac{2\sin x\cos x-\sin4x}{2\cos^22x-\cos2x}\\=\dfrac{\sin 2x-2\sin2x\cos2x}{\cos2x(2\cos2x-1)}\\=\dfrac{\sin 2x(1-2\cos2x)}{\cos2x( … Bình luận
Giải thích các bước giải: $VT=\dfrac{2\sin x\cos x-\sin4x}{2\cos^22x-\cos2x}\\=\dfrac{\sin 2x-2\sin2x\cos2x}{\cos2x(2\cos2x-1)}\\=\dfrac{\sin 2x(1-2\cos2x)}{\cos2x(2\cos2x-1)}\\=\dfrac{-\sin 2x(2\cos2x-1)}{\cos2x(2\cos2x-1)}\\=-\dfrac{\sin2x}{\cos2x}\\=-\tan2x\\=\tan(-2x)=VP\Rightarrow đpcm$ Bình luận
$VT=\dfrac{2\sin x\cos x-\sin4x}{2\cos^22x-\cos2x}\\
=\dfrac{\sin 2x-2\sin2x\cos2x}{\cos2x(2\cos2x-1)}\\
=\dfrac{\sin 2x(1-2\cos2x)}{\cos2x( …
Giải thích các bước giải:
$VT=\dfrac{2\sin x\cos x-\sin4x}{2\cos^22x-\cos2x}\\
=\dfrac{\sin 2x-2\sin2x\cos2x}{\cos2x(2\cos2x-1)}\\
=\dfrac{\sin 2x(1-2\cos2x)}{\cos2x(2\cos2x-1)}\\
=\dfrac{-\sin 2x(2\cos2x-1)}{\cos2x(2\cos2x-1)}\\
=-\dfrac{\sin2x}{\cos2x}\\
=-\tan2x\\
=\tan(-2x)=VP\Rightarrow đpcm$